How do you find the roots, real and imaginary, of $y=(x+4)(x+1)-3x^2+2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y=(x+4)(x+1)-3x^2+2$ using the quadratic formula?

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Answer 1 · 2016-07-14T12:26:51

$x = 5/4 +- (sqrt(73))/4$

Explanation:

Multiply out the bracket:

$(x+4)(x+1) = x^2 + 5x + 4$

$therefore y = -2x^2 + 5x + 6$

For quadratic $y = ax^2 + bx + c$ the roots occur at $y=0$ and are found with the quadratic formula. As it is a polynomial of order 2, we expect to get two roots.

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$x = (-5+-sqrt(25-4(-2)(6)))/(2(-2)) = (-5+-sqrt(73))/(-4)$

$therefore x = 5/4 +- (sqrt(73))/4$

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How do you find the roots, real and imaginary, of $y=(x+4)(x+1)-3x^2+2$ using the quadratic formula?

1 Answer

Explanation:

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